Answer:
[tex]9.495 \times 10^3\ m[/tex]
Explanation:
From the given information:
Using the equation of Barometric formula as related to density, we have:
[tex]\rho (z) = \rho (0) e^{(-\dfrac{z}{H})} \ \ \ \ --- (1)[/tex]
Here;
[tex]p(z) =[/tex] the gas density at altitude z
[tex]\rho(0) =[/tex] the gas density at sea level
H = height of the scale
[tex]H = \dfrac{RT}{M_ag } \ \ \ --- (2)[/tex]
Also;
R represent the gas constant
temperature (T) a= 280 K
g = gravity
[tex]M_a =[/tex] molaar mass of gas; here, the gas is Oxygen:
∴
[tex]M_a =[/tex] 15.99 g/mol
= 15.99 × 10⁻³ kg/mol
[tex]H = \dfrac{8.3144 \times 280}{15.99 \times 10^{-3} \times 9.8 }[/tex]
[tex]H =14856.43 \ m[/tex]
Now we need to figure out how far above sea level the density of oxygen drops to half of what it is at sea level.
This implies that we have to calculate z;
i.e. [tex]\rho(z) =\dfrac{\rho(0) }{(2)}[/tex]
By using the value of H and [tex]\rho(z)[/tex] from (1), we have:
[tex]\dfrac{\rho(0) }{(2)} = \rho (0) e^{(-\dfrac{z}{14856.43})}[/tex]
∴
[tex]\dfrac{1}{2} = e^{(-\dfrac{z}{14856.43})} \\ \\ e^{(-\dfrac{z}{14856.43})} =\dfrac{1}{2}[/tex]
By rearrangement and taking the logarithm of the above equation; we have:
[tex]- z = 14856.43 \times \mathtt{In}\dfrac{1}{2} \\ \\ -z = 14856.43 \times (-0.6391) \\ \\ z = 9495 \ m \\ \\ z = 9.495 \times 10^3\ m[/tex]
As a result, the oxygen density at [tex]9.495 \times 10^3\ m[/tex] is half of what it is at sea level.
7)
Which wave carries the most energy?
-)
A)
B)
C)
D)
Hover over answer image to enlarge
C.
Explanation:
C. is a gamma ray and they carry the most energy.
The most energy is carried by the gamma rays. The correct option is C.
What is a gamma ray?Gamma radiation, also known as gamma rays, is a type of electromagnetic radiation produced by the radioactive decay of atomic nuclei. It is made up of electromagnetic waves with the shortest wavelengths, which are typically shorter than those of X-rays.
Waves with a higher frequency have more energy. Gamma rays have the shortest wavelengths and highest frequencies of any electromagnetic wave. Gamma rays have more energy than any other electromagnetic wave due to their extremely high frequencies.
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Can someone please help me on this?
how do stocks and bonds differ
Answer:
Explanation:
Stocks give you partial ownership in a corporation, while bonds are a loan from you to a company or government. The biggest difference between them is how they generate profit: stocks must appreciate in value and be sold later on the stock market, while most bonds pay fixed interest over time.
Answer:
Stocks give you partial ownership in a corporation, while bonds are a loan from you to a company or government. The biggest difference between them is how they generate profit: stocks must appreciate in value and be sold later on the stock market, while most bonds pay fixed interest over time.
Explanation:
None
Which category best describes this group of shapes? rhombuses rectangles parallelograms squares II
they're parallelograms
Which electromagnetic wave has the shortest wavelength?
Which electromagnetic wave would have a wavelength approximately the size of the tip of a pencil?
Which electromagnetic wave has the longest wavelength?
Answer:
shortest - gamma Ray's
size of a pencil tip : most likely microwaves
longest - radio waves
Gamma rays have the shortest wavelength whereas Radio waves have the longest wavelength. The wave whose wavelength is about the size of the tip of a pencil is infrared ray.
What is an Electromagnetic spectrum?The whole electromagnetic waves are divided into several classes based on their frequencies or energies as follows, in the order of increasing frequency:
Radio waves
Microwaves
Infrared
Visible
Ultraviolet
X-rays
Gamma rays
The Gamma rays are the shortest wavelength group with frequencies greater than 3 × 10²⁰ Hz.
The Radio waves are the longest wavelength group. Their frequency is lower than 3 × 10⁷ Hz.
The tip of a pencil is about 0.5mm in width.
Calculate the frequency of the wave whose wavelength is about the size of the tip of a pencil i.e. λ = 0.5 mm,
ν = 3×10⁸/(0.5×10⁻³) = 6×10¹¹ Hz.
The infrared rays have frequencies in the range of 3×10¹¹ Hz to 3×10¹³ Hz. So, the wave is infrared.
Infrared waves have a wavelength about the tip of a pencil.
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1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
(D) x = 93.8 m
Explanation:
v^2 = v0^2 + 2ax
(20 m/s)^2 = (5 m/s)^2 + 2(2 m/s^2)x
Solving for x,
x = 93.8 m
A ball bounces changing velocity from vi=15m/s[D] to vf=15m/s[U] in t=0.01s. The balls acceleration is *
A) 0
B) 30m/s/s
C) 3000m/s/s
D) 0.3m/s/s
Answer:
Option (A) is correct
Explanation:
a= (vf-vi)/ t
put the values
hence,
a= ( 15-15)/0.01
a=0
This is energy of motion
Hint
First two letters Ki
Answer: Kinetic Energy
Explanation:Kinetic energy is the energy of motion. All moving objects have kinetic energy.
Answer:
Kinetic energy
Explanation:
Kinetic energy is energy produced when the object is in motion.
Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.
Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
Determine the Thrust developed
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : calculate the area of the duct
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
next : calculate the velocity of propeller
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
Finally determine the thrust developed
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids.An object is made from a uniform piece of sheet metal. The object has dimensions of a
This question is not complete, the complete question is;
A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids.
An object is made from a uniform piece of sheet metal. The object has dimensions of α = 1.50 ft, where α is the diameter the semi-circle, b= 3.51 ft, and c = 2.20 ft. A hole with diameter d = 0.500 ft is centered at ( 1.21, 0.750 ).
Find x", y", the coordinates of the body's centroid.
Answer:
x" = 1.4857 ft
y" = 0.668 ft
Explanation:
Given the data in the question and as illustrated in the second image below;
from the image;
BC² = DC² - BD²
BC² = 2.2² - 1.5² = 4.84 - 2.25 = 2.59
BC = √2.59 = 1.61 ft
AB = 3.51 ft - 0.75 ft - 1.61 ft = 1.15 ft
so;
A₁ = [tex]\frac{1}{2}[/tex] × 1.51 ft × 1.61 ft = 1.2075 ft²
x₁ = 0.75 + 1.15 + [tex]\frac{1}{3}[/tex](1.61 ft) = 2.44 ft
y₁ = [tex]\frac{1}{3}[/tex](1.5 ft) = 0.5 ft
A₂ = 1.15 ft × 1.5 ft = 1.725 ft²
x₂ = 0.75 ft + ( 1.15/2 )ft = 1.325 ft
y₂ = ( 1.5/2 ) ft = 0.75 ft
A₃ = [tex]\frac{\pi }{2}[/tex](0.75 ft)² = 0.88 ft²
x₃ = 0.75 - ([tex]\frac{4 }{3\pi }[/tex](0.75 ft)) = 0.43 ft
y₃ = 0.75 ft
diameter d = 0.5 ft and centered at ( 1.21, 0.750 )
A₄ = [tex]\frac{\pi }{4}[/tex]( d )² =
x₄ = 1.21 ft
y₄ = 0.75 ft
Thus;
x" = [tex]\frac{A_1 x_1 + A_2 x_2 + A_3 x_3 - A_4x_4 }{A_1+A_2+A_3-A_4}[/tex]
so we substitute
x" = [tex]\frac{(1.2075X2.44) + (1.725 X 1.325) + (0.88X0.43) - (0.196 X 1.21 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]
x" = [tex]\frac{ (2.9463 + 2.285625 + 0.3784 - 0.23716)}{ 3.6165 }[/tex]
x" = 5.373165 / 3.6165
x" = 1.4857 ft
y" = [tex]\frac{A_1 y_1 + A_2 y_2 + A_3 y_3 - A_4y_4 }{A_1+A_2+A_3-A_4}[/tex]
so we substitute
y" = [tex]\frac{(1.2075X0.5) + (1.725 X 0.75) + (0.88X0.75) - (0.196 X 0.75 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]
y" = [tex]\frac{ (0.60375 + 1.29375 + 0.66 - 0.14112)}{ 3.6165 }[/tex]
y" = 2.41638 / 3.6165
y" = 0.668 ft
Therefore,
x" = 1.4857 ft
y" = 0.668 ft
the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m , what is the springs displacement
Answer:
Spring's displacement, x = -0.04 meters.
Explanation:
Let the spring's displacement be x.
Given the following data;
Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg
Number of shrews, n = 49
Spring constant, k = 24 N/m
We know that acceleration due to gravity, g is equal to 9.8 m/s².
To find the spring's displacement;
At equilibrium position:
Fnet = Felastic + Fg = 0
But, Felastic = -kx
Total mass, Mt = nm
Fg = -Mt = -nmg
-kx -nmg = 0
Rearranging, we have;
kx = -nmg
Making x the subject of formula, we have;
[tex] x = \frac {-nmg}{k} [/tex]
Substituting into the formula, we have;
[tex] x = \frac {-49*0.002*9.8}{24} [/tex]
[tex] x = \frac {-0.9604}{24} [/tex]
x = -0.04 m
Therefore, the spring's displacement is -0.04 meters.
can someone please help me answer this!!
Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direction along the inner central wire, and the same current I runs in the opposite direction in the hollow conductor. How does the magnetic field B outside the entire cable (outside both the inner wire as well as the hollow conductor) vary as a function of distance away from the cable
Answer:
0 < r < r_exterior B_total = [tex]\frac{\mu_o I}{2\pi r}[/tex]
r > r_exterior B_total = 0
Explanation:
The magnetic field created by the wire can be found using Ampere's law
∫ B. ds = μ₀ I
bold indicates vectors and the current is inside the selected path
outside the inner cable
B₁ (2π r) = μ₀ I
B₁ = [tex]\frac{\mu_o I}{2\pi r}[/tex]
the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.
For the outer shell
for the case r> r_exterior
B₂ = \frac{\mu_o I}{2\pi r}
This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction
for the case r <r_exterior
in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero
B₂ = 0
Now we can find the field created by each part
0 < r < r_exterior
B_total = B₁
B_total = [tex]\frac{\mu_o I}{2\pi r}[/tex]
r > r_exterior
B_total = B₁ -B₂
B_total = 0
A negative charge of 2 C and a positive charge of 3 C are separated by 80 m. What is the force between the two charges?
Answer:
6
Explanation: i looked this up.
According to the Coloumb's law, the force is obtained from the charges and the distance between them is 8.4375×10⁶ N. This force results in the attractive force.
What is Coloumb's law?Coloumb's law states the relation between the charges and the distance between the charges. Coloumb's law states that the Force is directly proportional to the product of charges and inversely proprotional to the square of distance between them.
The Coloumb's law gives the force of attraction or repulsion between the charged bodies. Two charges with positive or negative charges repels each other. One positive and one negative charge attract each other.
The force increases with the product of charges increases as product of charges and force are directly proprotional. The force decreases with the increase in distance of seperation and vice-versa. The SI unit of force is newton (N).
The Coloumb's law is:
F = k (q₁×q₂ / r²)
k is the constant of proportionality and is equal to 9×10⁹ N.m²/C².
q₁, q₂ = charges
r² = distance of seperation of charges.
From the given,
q₁ = -2C ( negative sign represents the negative charge)
q₂ = +3C ( + sign represents the positive charge)
r² = 80 m ( distance between the charges)
F = k (q₁×q₂ / r²)
= 9×10⁹×2×3 / 80×80
= 54×10⁹ / 6400
= 8437500 N
F = 8.4375×10⁶ N
There is the presesnce of both positive and negative charges, hence it results in the attractive force. Hence, the force F between two charges separated by 80m is 8.4375×10⁶ N.
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What is the flux through the plane taken parnllel to the field?
Answer:
the flux through the plane is zero.
Explanation:
You and your friends are pushing a car with a total of 1080 N of force. You are able to push the car for a distance of 218 m. Calculate the work done on the car.
Answer:
How much power is required to pull a sled if you use 60J of work in 5 seconds? ... Misfortune occurs and Renatta and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push their car 218 m to the ... Calculate the amount of work done when moving a 567N crate a distance of 20 meters.
Explanation:
Misfortune occurs and Renatta and her friends find themselves getting a ... They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel ... Write down what they give you. ... Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m ... Work = Force x Distance = Joules (force = weight).
A particle oscillates in simple harmonic motion, with amplitude A and period T. The particle starts from position x = A. What is the magnitude of the particle's displacement a time 3T/2?
A. 3A
B. 6A
C. 2A
D. 0
Answer:
i guess it is D because,
y = A sin(wt)
w = [tex]\frac{2\pi }{T}[/tex] and t = 3T/ 2
now, Y = A sin ( [tex]\frac{2\pi }{T} X \frac{3T}{2}[/tex])
so, Y = 0radian
Explanation:
The magnitude of the particle's displacement a time 3T/2 is zero. The correct option is D.
What is magnitude?Magnitude is simply "distance or quantity" in the context of physics.
A = amplitude of SHM
T = time period of SHM
At t = 0 s, the position of a particle is, x = A
At t = 3T/2, find the magnitude of the particle's displacement
d = magnitude of displacement of particle at time
t = 3T/2
t = ?
t= t1 + t2 = T + T/2
where
t1 = T
t2 = T/2
Now, in time t1 = T, the particle will return to its initial position as the particle completes one full cycle of motion in time T.
Therefore, after time t1 = T, the position of the particle is given by x = A, and the total displacement by the particle is 0.
Now in time
t2=T/2
The particle will go to another extreme position, given by x = -A.
Therefore, the net displacement of particles in time
3/T2
is given by:
d=0+(−A)=−A
d=−A
Therefore, the correct option is D. 0.
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A marble and a basketball are rolling toward you at the same velocity. If the same amount of force is used to stop both objects, what is true about the amount of time needed to stop those objects?
Answer:
The amount of time needed to stop the marble is much less than the amount of time needed to stop the basketball.
Explanation:
To explain this question we can use the equation:
Change in momentum = Force x Time
Where momentum = mass x Velocity
In the case of the basketball, it has a much larger mass than the marble, meaning it has much more momentum than the marble. This means that with the same amount of force, it will need much more time to stop.
Likewise, the marble has less mass than the basketball meaning it will have less momentum. This means that with the same amount of force, it will require less time to stop.
Logically this also makes sense if we think about it in a real life scenario. If we try to stop a marble with our finger, we can stop it much faster then if we try to stop a basketball with our finger because of the difference in mass between the two objects, therefore the marble will require much less time to be stopped than the basketball.
Hope this helped!
The first extra-solar planet around a main sequence star (i.e., not a neutron star or white dwarf) was found around the star 51 Pegasi in 1995. The large planet causes a measureable motion of the star around the center of mass of the system. Pegasi 51's orbital motion had a period of 4 days indicating a very large planet very close to the star. A large planet of mass M around a small star of mass 4M. The distance between planet and star is L. Both the planet and star will orbit around the center of mass of the system (marked by the red X).
Required:
Where is the center of mass of the star-planet system?
Answer:
[tex]r_{cm}[/tex]= 1/5 L
Explanation:
To find the center of mass of the system let's use
[tex]r_{cm} = \frac{1}{M}[/tex] ∑ r_i x_i
where m is the total mass of the system
let's apply this expression to our case
Let's set the reference frame on the star
[tex]r_{cm} = \frac{1}{M +4M} ( 4M 0 + M L)[/tex]
r_{cm} = [tex]\frac{1}{5}[/tex] L
[tex]r_{cm}[/tex]= 1/5 L
Hey, the breast center is 1/5 of the distance between the star and the planet.
8. What is the mass and charge of a proton?
A. 1 amu, no charge
B. 1 amu, charge of +1
C. Mass of O, charge of -1
D. None of the above
Two risks of exposure to High levels of UV radiation
Answer:
uv radiation cause cancer
uv radiation effect our eyes
Answer:
you can get
1:skin cancer
2:eye damage
3:skin damage
4:immune system suppression
choose which two u want
hope this helped
:)
Explanation:
An Object, Start from rest w Confront Aiceleration 8m/s2 along a
Straight line. Find
A, the speed At the end Of 5 second
B, The average Speed for the 5second interval
Answer:
A) v = 40 m / s, B) v_average = 20 m / s
Explanation:
For this exercise we will use the kinematics relations
A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero
v = vo + a t
v = 0 + 8 5
v = 40 m / s
B) the average velocity can be found with the relation
v_average = vf + vo / 2
v-average = 0+ 40/2
v_average = 20 m / s
Average running speed is 8km/h. Convert this to m/s :
a)8m/s
b)2.2m/s
c)28.8m/s
d)8km/h
Answer:
Im not sure if you mean miles per hour or meters per hour but if it is miles it is 4.97 miles per hour and if it is meters then it is 8000 meters per hour
Explanation:
Sorry if I'm wrong
You are riding on a carousel that is rotating at a constant 24 rpm. It has an inside radius of 4 ftand outside radius of 12 ft. You begin to run from the inside to the outside along a radius. Your peak velocity with respect to the carousel is 6 mph and occurs at a radius of 8 ft.What are your maximum Coriolis acceleration magnitude and its directionwith respect to the carousel
Answer:
magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
Explanation:
Given the data in the question;
Speed of carousel N = 24 rpm
From the diagram below, selected path direction defines the Axis of slip.
Hence, The Coriolis is acting along the axis of transmission
Now, we determine the angular speed ω of the carousel.
ω = 2πN / 60
we substitute in the value of N
ω = (2π × 24) / 60
ω = 2.5133 rad/s
Next, we convert the given velocity from mph to ft/s
we know that; 1 mph = 1.4667 ft/s
so
[tex]V_{slip[/tex] = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s
Now, we determine the magnitude of the Coriolis acceleration
[tex]a_c[/tex] = 2( [tex]V_{slip[/tex] × ω )
we substitute
[tex]a_c[/tex] = 2( 8.8002 ft/s × 2.5133 rad/s )
[tex]a_c[/tex] = 44.235 ft/s²
Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
give 3 example of right quantities that have no unit
Answer:
velocity ratio
mechanical advantage
Relative density
A bucket of water of mass 14.6 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.320 m with mass 11.1 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 11.0 m to the water. You can ignore the weight of the rope.
Required:
a. What is the tension in the rope while the bucket is falling?
b. What is the time of fall?
c. While the bucket is falling, what is the force exerted on the cylinder by the axle?
Answer:
a. T = 39.41 N
b. t = 1.76s
c. 150.78 N
Explanation:
Given:
Mass of bucket of water, Mb = 14.6 kg
Mass of cylinder, Mc = 11.1 kg
Diameter of cylinder, D = 0.320 m, or radius, r = D/2 = 0.16m
Displacement of the bucket from the top, that is the vertical displacement , y = 11.0 m
a. The tension in the rope while the bucket is falling is:
F = mg - T = ma
Where F= The force
m= mass
g= Acceleration due to gravity
T = tension in the rope
a = acceleration
T= m(g - a)
Then, calculating the angular acceleration of the pulley system in relation to its radial acceleration
T= 1/2Ma
Merging the two final equation so as to solve for a
M(g - a) = 1/2Ma
Make a the subject of the formula
Mg - Ma = 1/2Ma
1/2Ma + Ma = Mg
a (1/2 M + M) = Mg
Divide both side by (1/2 M + M)
a = Mg ÷ (1/2 M + M)
Inputing the given value in the formula above
g= 9.8m/s2
a = (14.6 kg) (9.8m/s2) ÷ 1/2 (11.1 kg) + 14.6 kg
a = 7.1007m/s2
Now it is easy to input the value into T= 1/2Ma
T = 1/2 (11.1 kg) (7.1007m/s2) = 39.41 N
B. Time of fall is:
Using one of the equation of motion
s = ut + 1/2 at^2
U = Initial velocity
t = time
a = acceleration
s= distance in this case displacement y
making t the subject of the formula
t = √(2s ÷ a)
u is 0 since the bucket starts from rest
so, t = √((2)(11.0 m) ÷ 7.1007m/s2)
t = 1.76s
c. the force exerted on the cylinder by the axle = T + Mg
= 42 N + (11.1 kg) (9.8m/s2)
= 150.78 N
The universe cooled after the Big Bang.At some point hydrogen atoms combined to form helium.What is this process called?
Answer:
Nuclear fusion
Explanation:
Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes. The data are shown in the table. Estimate the average temperature of the air in the room at 20 min. explain your answer.
The average temperature of the air in the room at 20 min is 23°C.
What is temperature?Temperature is the degree of hotness or coldness of the object.
A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes.
The surrounding is vast, its temperature does not get affected by small amount of water. So, the temperature of air remains constant.
Thus, the average temperature of the air in the room at 20 min is 23°C.
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1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
An object is pushed with a force of 400 N for 20 meters; what is the work done on the object? Solve using W=F*d
Answer:
800joules
Explanation:
work is measured in joules